Question

A circle is shown. 2 raddi are drawn. A chord connects the points of the radii on the circle to each other to form a triangle. The radii have lengths of 5 feet. A perpendicular bisector cuts through an angle with a measure of 72 degrees and cuts the opposite side of the triangle into 2 equals parts with measures of 2.9 feet. The length of the bisector is 4 feet. The space between the chord and the outline of the circle is shaded. What is the area of the shaded portion of the circle? (5π – 11.6) ft2 (5π – 5.8) ft2 (25π – 11.6) ft2 (25π – 5.8) ft2

Answer 1

Answer:

$(5\pi-11.6)ft^2$

Step-by-step explanation:

Given:

Length of radius of circle = 5 feets

Length of perpendicular bisector = 4 feets

To find:

Area of the shaded portion of the circle

Solution:

As OD is perpendicular bisector of AB,

$AB=2AD=2(2.9)=5.8\,\,feets$

$\angle ODA=90^{\circ}$

Area of $\Delta AOB$ = 1/2 (base) × (height) = $\frac{1}{2}\times 4\times 5.8=11.6$  square feets

Area of sector AOBC = $\frac{\angle AOB}{360^{\circ}}\pi(r^2)=\frac{72^{\circ}}{360^{\circ}}\pi(5^2)=5\pi$  square feets

Here, r denotes radius of circle

So,

Area of shaded portion = Area of sector AOBC - Area of $\Delta AOB$ = $(5\pi-11.6)ft^2$  

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